Return to the credit card scenario of exercise 12 (section 2.2), and let c be the event that the selected student has an american express card. in addition to p(a) = .6, p(b) = .4, and p(a > b.= .3, suppose that p(c) = .2, p(a > c) = .15, p(b >
c.= .1, and p(a > b >
c.= .08.
a. what is the probability that the selected student has at least one of the three types of cards?
b. what is the probability that the selected student has both a visa card and a mastercard but not an american express card?
c. calculate and interpret p(b u
a. and also p(a u b).

fa tha first one you gonna add y to both sides

it look like this

x-y+y=6+y

x=y+6

therefore your answer is

x =y+6

fa tha second one

you gonna flip tha equation

it gonna look like x-4y=y

then add 4 to both sides

it gonna look like tha

x-4 +4 =y +4

so therefore tha answer is x=y+4

step-by-step explanation:

step-by-step explanation:

change of stock price/initial stock price on friday * total stock value = change of stock value

0.89/73.73 *x = 42.72

x=3539.04 = total stock value at initial stock price on friday .

total stock value at initial stock price on friday/ initial stock price on friday = number of shares

3539.04/73.73 = 48 shares

they own 48 shares of this stock

if i'm not wrong, i would say that the third angle has (72) because the three sides of all triangles have 180 degrees and if we subtract 180 - 62 - 40 = (72) / but i'm not sure about 40 yd.